\documentclass[11pt]{article}

\usepackage{precalc}

\begin{document}

\assigntitle{15}{Vectors}

\topic{rectangular/polar duality}
Last week, you studied the duality between \term{rectangular} and
\term{polar} coordinates; as you learned, they are two different ways
of representing the same information (position in two-dimensional
space), and we can easily convert back and forth between the two
representations.

But why would we care that there are two different representations?
Can't we just pick one and use that?  The answer, of course, is that
each representation is useful for different things, and is related to
other mathematical concepts in different ways.

This rectangular/polar duality comes up in two other particular places
we're going to study.  The first, vectors, are a way of formalizing the
ideas of motion and force.  The second may be unexpected: complex
numbers also exhibit this rectangular/polar duality!  If you've seen
complex numbers before, you probably learned only about the
rectangular representation: $a + bi$.  But complex numbers can also be
thought of in polar terms, which we will study in an upcoming
assignment.

\section{The Racecar Game}
\label{sec:racecar}

\topic{vroom!}
Before starting this assignment (or at least at some point during the
week), try playing the racecar game, which is described in a second
document linked from the webpage!

\section{Vectors as Arrows}
\label{sec:vectors-arrows}

\topic{vector definition}
A \term{vector} is a mathematical object that represents an
\emph{amount} (usually called the \term{magnitude}) as well as a
\emph{direction}.  For example, each segment that a car traveled in
the racecar game was a vector: it had an amount (how far the car
traveled) and a direction.  A force can also be represented as a
vector: it has an amount (the strength of the force) as well as a
direction (the direction in which the force is pushing).

Graphically, we can represent vectors using arrows.  The \emph{length}
of the arrow represents the magnitude of the vector, and the direction
of the arrow represents the direction of the vector.  The end with the
arrowhead is called the \term{head}, and the other end is called the
\term{tail}.

\diagram{vector}{A vector.}

Since vectors just have a direction and a magnitude, it doesn't matter
where a vector starts or ends.  Moving a vector around doesn't change
it at all.  \pref{fig:parallel-vectors} shows three vectors which are
all equal.

\diagram{parallel-vectors}{Three equal vectors.}

\section{Vector Addition}
\label{sec:vec-addition}

We can add two vectors together to produce another vector, and this
means exactly what you might think.  If you travel 40 feet in a
certain direction, and then 50 feet in another direction, what how far
and in what direction did you travel altogether?  If two people are
pushing on a box in different directions and with different strengths,
in what direction will the box move?

\topic{vector addition graphically}
Graphically, vector addition can be represented by putting the two
vectors to be added head-to-tail; then their sum is the vector from
the head of the first to the tail of the second, as shown in
\pref{fig:vec-addition}.  We put a little arrow over a variable (like
$\vec{a}$) to indicate that we are talking about a vector.  To make
something like $\vec{a}$ in \LaTeX, you can type \verb|\vec{a}|.

\diagram{vec-addition}{Adding the vectors $\vec{a}$ and $\vec{b}$.}

We can use the Law of Cosines and Law of Sines to compute the sum of
two vectors.

\diagram{vec-add-example}{Adding two vectors using the Law of Cosines}

\begin{problem} \label{prob:loc-sum}
  Consider \pref{fig:vec-add-example}, which shows two vectors, one
  with length $12$ and angle $50^\circ$, and the other with length $5$
  and angle $-30^\circ$ (as usual, we measure vector direction as
  angles from the positive x-direction).  If we add these two vectors,
  we get the third vector shown on the bottom.  Let's compute the
  magnitude and direction of this vector.
  \begin{subproblems}
    \item First, find angle $B$.
    \item Now use the Law of Cosines to find the length of $AC$.
    \item Use the Law of Sines to find angle $BAC$.
    \item Finally, use this to find the direction of vector $AC$
      (marked $?^\circ$ in the diagram).
  \end{subproblems}
\end{problem}

\section{Vectors and rectangular coordinates}
\label{sec:vectors-rect}

Adding vectors using the Law of Cosines in this way is rather
cumbersome.  Isn't there a better way?  As it turns out, there is!

A vector has a \term{magnitude} and a \term{direction}\dots a
\emph{length} and an \emph{angle}\dots doesn't this sound familiar?
It's just like polar coordinates!  In fact, it \emph{is} polar
coordinates.  So maybe there's a way to represent vectors using
rectangular coordinates, too?

\topic{vector decomposition}
Sure enough, there is!  Any vector $\vec{v}$ can be decomposed into
two vectors, one vertical and one horizontal, whose vector sum is
$\vec{v}$.  An example is shown in \pref{fig:vec-decomp}.

\diagram{vec-decomp}{Decomposing $\vec{v}$ as $\vec{x} + \vec{y}$}

But you already know how to do this from last week's assignment!  It's
just converting from polar to rectangular coordinates.

\begin{problem}
  Convert vector $AB$ in \pref{fig:vec-add-example} to rectangular
  coordinates.
\end{problem}

\begin{problem}
  Convert vector $BC$ in \pref{fig:vec-add-example} to rectangular
  coordinates.
\end{problem}

The nice thing is that now the vectors are very easy to add: just add
the $x$-components, and add the $y$-components.

\begin{problem} \label{prob:rect-sum}
  Add the vectors from the two previous problems using rectangular
  coordinates.
\end{problem}

Now we can convert back to polar coordinates: you already know how to
do this, too.

\begin{problem}
  Convert the vector from \pref{prob:rect-sum} back into polar
  coordinates.  Does it agree with your answers to \pref{prob:loc-sum}?
\end{problem}

\begin{problem}
  \topic{a Hard(ing) problem}
  Adler, Bartholomew, and Custer are pushing a large marble statue of
  President Harding on wheels.  Adler is pushing at an angle of
  $20^\circ$ with a force of $50$ Newtons.  Bartholomew is pushing at
  an angle of $25^\circ$ with a force of $40$ Newtons.  Custer doesn't
  want them to move the statue, so he is pushing back at an angle of
  $185^\circ$ with a force of $60$ Newtons.  In what direction and
  with what force does the statue move? Show your work, and express
  your answers rounded to the nearest three decimal places.

  (Hint: to find the total force you can just add the vectors
  corresponding to each person's pushing.  You can either use the Law
  of Cosines (twice!) or (recommended) convert to rectangular
  coordinates first, add component-wise, and convert back to polar
  coordinates.)

\diagram{statue}{A statue-pushing match}
\end{problem}

\end{document}